A beam of electrons of energy E scatters from a target having atomic spacing of 1 mathringA The first maximum intensity occurs at θ=60° Then E (in eV ) is . (Planck constant h =6.64 × 10-34 Js 1 eV =1.6 × 10-19 J , electron mass . m =9.1 × 10-31 kg )

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A beam of electrons of energy E scatters from a target having atomic spacing of 1 mathringA The first maximum intensity occurs at θ=60° Then E (in eV ) is . (Planck constant h =6.64 × 10-34 Js 1 eV =1.6 × 10-19 J , electron mass . m =9.1 × 10-31 kg ) (A)  (B)  (C)  (D)

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<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/9d460be943499f19400c5ea672ad896a-.png" /> 2 d sin θ=λ=( h /√2 mE ) 2 × 10-10 × (√3/2)=(6.6 × 10-34/√2 mE ) E=(1/2) × (6.642 × 10-43/9.1 × 10-31 × 3 × 1.6 × 10-19)=50.47

Q. A beam of electrons of energy E scatters from a target having atomic spacing of 1A˚ The first maximum intensity occurs at θ=60∘ Then E (in eV ) is ____. (Planck constant h=6.64×10−34Js 1eV=1.6×10−19J, electron mass m=9.1×10−31kg)

2dsinθ=λ=2mE​h​ 2×10−10×23​​=2mE​6.6×10−34​ E=21​×9.1×10−31×3×1.6×10−196.642×10−43​=50.47

$2 d sin θ = λ = \frac{h}{2 m E}$
$2 \times 1 0^{- 10} \times \frac{3}{2} = \frac{6.6 \times 1 0 ^{- 34}}{2 m E}$
$E = \frac{1}{2} \times \frac{6.6 4 ^{2} \times 1 0 ^{- 43}}{9.1 \times 1 0 ^{- 31} \times 3 \times 1.6 \times 1 0 ^{- 19}} = 50.47$